Two different limits for two similar polar coordinates expressions

Question

I have the following two multivariable function limits:

$\displaystyle \lim _{(x, y) \rightarrow(0,0)} \frac{x y^{3}}{x^{2}+y^{6}}:f_{1}$ and $\displaystyle \lim _{(x, y) \rightarrow(0,0)} \frac{x^{2} y^{2}}{x^{2}+y^{6}}: f_{2}$

I have concluded, by using traditional methods (Approaching from different lines, and the precise definition of limit), that the first limit is DNE, while the second one is $0$

However, I was trying to use polar coordinates to prove. They have very similar expressions.

From this post, I understand that the function has to be bounded in order for the limit to be 0.

$\displaystyle f_1\rightarrow \lim _{(x, y)\rightarrow(0,0)} \frac{r \cos \theta \cdot r^{3} \sin ^{3} \theta}{r^{2} \cos ^{2} \theta+r^{6} \sin^{6} \theta} =\frac{r^{2} \sin ^{3} \theta \cos \theta}{\cos ^{2} \theta+r^{4} \sin^{6} \theta}$

$\displaystyle f_2\rightarrow \lim _{(x, y)\rightarrow(0,0)} \frac{r^{2} \cos ^{2} \theta \cdot r^{2} \sin ^{2} \theta}{r^{2} \cos ^{2} \theta+r^{6} \sin^{6} \theta} = \frac{r^{2} \sin ^{2} \theta \cos^{2} \theta}{\cos ^{2} \theta+r^{4} \sin^{6} \theta} $

However, to me, by having "$\displaystyle r^{4} \sin^{6} \theta$" in the denominator both functions are not bounded. I don't know how two different numerators in this case change the situation?

Is there any good reason to explain this?

Thanks ! In this way, for $f_1$ it becomes a function that is free of r. As a result, it will change depends on the value of $\theta$. However, what about the second one? — Yan Zhuang, Jul 23 '20 at 04:57

score 1 · Answer 1 · answered Jul 25 '20 at 11:18

1

If you neglect the term $r^4\sin^6\theta$ when $r$ gets small, the two fractions simplify to

$$\frac{r^2\sin^3\theta}{\cos\theta}$$ and $$r^2\sin^2\theta.$$ The first one is unbounded. The Cartesian equivalents are (for small $y$)

$$\frac{y^3}x$$ and $$y^2.$$

answered Jul 25 '20 at 11:18

Thank you soooo much! But why we are about to ignore the term $\displaystyle r^4 sin^6\theta $? It won't affect our result? – Yan Zhuang Jul 26 '20 at 15:52
I understand the fact that the polar coordinate has to be bounded for the limit to equal to the zero. But with still $\displaystyle r^4 sin^6\theta $ left in the denominator, it made me hard to figure it out. – Yan Zhuang Jul 26 '20 at 15:54

score 0 · Answer 2 · edited Jul 25 '20 at 11:08

0

For the second one we can use that by AM-GM

$$x^{2}+y^{6}=\frac{x^2}2+\frac{x^2}2+y^6 \ge \frac{3}{\sqrt[3]{4}}|x|^\frac4{3}y^2$$

and then

$$0\le \frac{x^{2} y^{2}}{x^{2}+y^{6}}\le \frac{\sqrt[3]{4}}{3}|x|^\frac 2 3 \to 0.$$

edited Jul 25 '20 at 11:08

Vincenzo Tibullo

11,287

answered Jul 23 '20 at 20:50

user

162,563

Two different limits for two similar polar coordinates expressions

2 Answers2